Question

The graph of which of the following equations has y=1 as an asymptote?

Answer 1

The answer is B.

To solve this, we replace y = 1 on each of the options. If there is a vertical asymptote, we should get an impossible result.

`1=(x^2)/(x-1)`

Now we solve for x, and we get a cuadratic expresion

`\begin{gathered} x-1=x^2 \\ 0=x^2-x+1 \end{gathered}`

Now we use the formula for the root of a quadratic expression:

`x_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Then replace all:

`x_(1,2)=\frac{1\pm\sqrt[]{1-4\cdot1\cdot1}}{2\cdot1}`

Before solve all this, we can look at the square root. Inside there is what is called the discriminant, this give us information about the roots:

if the discriminant is positive, the function has two real root

If the discriminant is equal to 0, we have two real roots, but they are equal

If the discriminant is negative, the function has two complex roots, thus doesn't has real roots.

In this case, 1-4 = -3. The function doesn't have real roots, so there is no possible value of x that satisfies y = 1

With this, we can say that there is a vertical asymptote on y = 1